Calculating Partial Derivative of Log(x^2+y^2) w/r/t x

[Somefantastik]

I'm trying to get

\frac{\partial}{\partial x} log(x^{2} + y^{2})

let z = x²+y²

Do I need to do a change of base to go from log₁₀z to log_ez before I can do the partial w.r.t. x?

That would make it

\frac{\partial}{\partial x} log(x^{2} + y^{2}) = \frac{1}{x^{2}+y^{2}} \ (2x) \ (log_{10}e)

Does this look right?

Then

\frac{\partial^{2}}{\partial x^{2}}(log(x^{2}+y^{2})) = - \frac{1.72x^{2}}{(x^{2} + y^{2})^{2}} + 0.86(x^{2}+y^{2})

?? that doesn't look right.

 

[tiny-tim]

Do I need to do a change of base to go from log₁₀z to log_ez before I can do the partial w.r.t. x?

That would make it

\frac{\partial}{\partial x} log(x^{2} + y^{2}) = \frac{1}{x^{2}+y^{2}} \ (2x) \ (log_{10}e)

Hi Somefantastik!

Yes, that's fine …

log_ab = \frac{log_eb}{log_a}

and of course log_ab = 1/log_ba

\frac{\partial^{2}}{\partial x^{2}}(log(x^{2}+y^{2})) = - \frac{1.72x^{2}}{(x^{2} + y^{2})^{2}} + 0.86(x^{2}+y^{2})

?? that doesn't look right.

ha ha!

how about \frac{0.86}{(x^{2}+y^{2})} ?
​

(and then of course combine both fractions into one)

 

[Somefantastik]

Hello tiny-tim, from the sunny south [usa] :)

That gives -0.86 /(x² + y²)

It just seems like a weird number to me.

That's going to be the same number for the partial with respect to y, isn't it?

They should cancel out; I'm trying to show that this is a harmonic using the laplacian.

 

[tiny-tim]

Hi Somefantastik!

That's going to be the same number for the partial with respect to y, isn't it?

They should cancel out; I'm trying to show that this is a harmonic using the laplacian.

Yup …

\left(\frac{\partial^{2}}{\partial x^{2}}\ +\ \frac{\partial^{2}}{\partial y^{2}}\right)(log(x^{2}+y^{2})) = - \frac{1.72(x^{2}\,+\,y^2)}{(x^{2} + y^{2})^{2}}\ +\ 0.86/(x^{2}+y^{2})\ +\ + 0.86/(x^{2}+y^{2})\ =\ 0\